As the given number number is a 16 digit binary number, so when chunk in 4 digit groups, it will form 4 groups. Each group will represent a single digit in hexadecimal code.

So, a 16-digit binary pattern converts to a 4 digit hexadecimal value.

7 0

3 years ago

A line passes through the points (–1, –5) and (4, 5). The point (a, 1) is also on the line.

IgorC [24]

Answer:

a = 2

Step-by-step explanation:

Let's find the equation of the line using the 2 points given.

Let's call -1 as x_1 and -5 as y_1

also, 4 as x_2 and 5 as y_2

The equation of line is :

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

Let's plug the points and get the equation of the line:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\y+5=\frac{5+5}{4+1}(x+1)\\y+5=2(x+1)\\y+5=2x+2\\y=2x-3$

Now, to find a, we substitute a in x and 1 in y of the equation of the line we just got:

$y=2x-3\\1=2(a)-3\\1=2a-3\\2a=4\\a=\frac{4}{2}=2$

4 0

3 years ago

Read 2 more answers

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago